Ideal Gas Mixtures
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Transcript Ideal Gas Mixtures
EGR 334 Thermodynamics
Chapter 12: Sections 1-4
Lecture 38:
Ideal Gas Mixtures
Quiz Today?
Today’s main concepts:
• Be able to describe ideal gas mixture composition in terms of
mass fractions and mole fractions.
• explain use of the Dalton model to relate pressure, volume,
and temperature and to calculate changes in U, H, and S for
ideal gas mixtures.
• Be able to apply mass, energy, and entropy balances to
systems involving ideal gas mixtures, including mixing
processes.
Reading Assignment:
Read Chapter 12, Sections 5-8
Homework Assignment:
Problems from Chap 12: 4, 10, 22, 28
Sec 12.1 : Describing Mixture Composition
3
Thus far we have been working with pure fluids
• Nitrogen
• Water
• Oxygen
• Air
• Ammonia
• R22
For gas mixtures, in addition to the normal to parameters (T, p), we
also need to know the mixture composition.
Number of moles
ni
mi
M
i
mass of i
Molecular
Weight
Mass fraction mf m i mass of i
i
mT
total mass
where
Mole fraction y n i moles of i
i
nT
total moles
where
Avg. Molecular Weight
M
mT
nT
of i
1
mf
i
i
to tal m ass
to tal m o les
1
nM
i
i
nT
yi
i
i
i
yi M
i
Sec 12.2 : Relating P, V, and T for Ideal Gas Mixtures
Ideal gas
pV nRT
4
This refers to the overall mixture, on average.
How do we describe the relationship between
p, V, and T for each component?
Dalton Model: Assumes that each component behaves as an ideal
gas at the specified T and V of the mixture. This assumes that the
gases do not interact with each other.
Partial Pressure
and
pi
p
pi
ni R T
V
ni R T V
nT R T V
ni
n
yi
thus
pi yi p
Amagat Model: Assumes that each component behaves as an ideal
gas at the specified T and p of the mixture.
Partial Volume
Vi
ni R T
p
and
V i y iV
5
Example (12.6): Natural gas at 23°C, 1 bar enters a furnace with the
following molar analysis:
40% propane (C3H8), 40% ethane (C2H6), and 20% methane (CH4).
Determine
(a) The analysis in terms of mass fractions
(b) The partial pressure of each component, in bar
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
i
C3H8
C2H6
CH4
Total
T (°C)
23
p (bar)
1
yi
Mi
mi
mfi
mi (kg/s)
0.40
0.40
0.20
1.0
yi is the mole fraction of
each component
6
Example (12.6):
40% propane (C3H8),
40% ethane (C2H6), and 20% methane (CH4).
(a) The analysis in terms of mass fractions
(b) The partial pressure of each component, in bar
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
Molecular weights of each component are found in Table A-1:
i
C33H88
C22H66
CH44
Total
T (°C)
23
p (bar)
1
yii
0.40
0.40
0.20
Mii
44.09
30.07
16.04
mii
17.636
12.028
3.208
1.0
32.872
mfii
mii(kg/s)
Units are effectively kg/kmol, thus mt = M
The mass of each component is
found
m y M
i
i
i
On a 1 kmol basis then
m P 0 . 4 44 . 09 17 . 636
m E 0 . 4 30 . 09 12 . 028
m M 0 . 2 16 . 04 3 . 208
m to ta l
m
i
i
3 2 .8 7 2
7
Example (12.6):
(b) The partial pressure of each component, in bar
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
To find Partial Pressure:
pi yi p
Total
p P 0 . 4 1bar
0 . 4 bar
T (°C)
23
1
0 . 4 bar
p (bar)
p E 0 . 4 1bar
i
C3H8
C2H6
CH4
yi
0.40
0.40
0.20
Mi
44.09
30.07
16.04
mi
17.636
12.028
3.208
32.872
mfi
0.5365
0.3659
0.0976
1.0
mi(kg/s)
1.0
p M 0 . 2 1bar
0 . 2 bar
8
Example (12.6):
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
First convert volumetric
flow rate to mass flow rate:
m
AV
AV P
v
RT
3
m
i
C3H8
C2H6
CH4
T (°C)
R
2
M T
2
( 2 0 m / s )(1 0 kN / m )
8 .3 1 4 kJ / km o l K
296 K
3 2 .8 7 2 kg / km o l
Total
23
p (bar)
0.40
0.40
0.20
1.0
yi
0.40
0.40
0.20
1.0
Mi
44.09
30.07
16.04
mi
17.636
12.028
3.208
32.872
mfi
0.5365
0.3659
0.0976
1.0
mi(kg/s)
AV p
26.71
kg
s
kJ
kN m
9
Example (12.6):
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
Can also find the mass flow rate of each component
mi
m fi
m
m P 0 .5 3 6 5 2 6 .7 1 kg / s
i
C3H8
C2H6
CH4
T (°C)
Total
23
P (bar)
0.40
0.40
0.20
1.0
yi
0.40
0.40
0.20
1.0
Mi
44.09
30.07
16.04
mi
17.636
12.028
3.208
32.872
mfi
0.5365
0.3659
0.0976
1.0
mi(kg/s)
14.33
9.77
2.61
26.71
14.33
kg
s
m E 0 .3 6 5 9 2 6 .7 1 kg / s
9.77
kg
s
m M 0 .0 9 7 6 2 6 .7 1 kg / s
2.61
kg
s
Sec 12.3 : Evaluating U, H, S and specific heats
10
The properties of U, H, S, and c of a mixture are additive.
Mass of system: m m m m ......
T
1
2
3
m
i
i
Internal energy
U T U 1 U 2 U 3 ......
U
i
i
mu
...... n u
...... y u
mu T m 1u 1 m 2 u 2 m 3 u 3 ......
n u T n1u 1 n 2 u 2 n 3 u 3
u T y 1u 1 y 2 u 2 y 3 u 3
i
i
i
i
i
i
i
i
i
Where a similar set of equations can be written for H and S
For specific heat c P y1 c P ,1 y 2 c P , 2 y 3 c P , 3 ......
cV y1 cV ,1 y 2 cV , 2 y 3 cV , 3 ......
y i c P ,i
i
i
y i cV , i
Sec 12.4 : Analyzing Systems Involving Mixtures
11
For mixtures with constant
composition (no chemical reaction):
U 2 U1
n u @ T u @ T
i
i
2
i
1
i
u
y u @ T u @ T
i
i
2
i
1
i
For constant specific heats,
u c v T 2 T1
with
u i c v , i T 2 T1
Equations for enthalpy (H) are similar to those for internal energy (U), but
uses cp.
Sec 12.4 : Analyzing Systems Involving Mixtures
12
For entropy, is also dependent upon pressure changes.
S 2 S1
n s @ T
i
i
s
i
2
, p 2 s i @ T1 , p1
y i s i @ T 2 , p i , 2 s i @ T1 , p i ,1
i
pi ,2
s s i @ T 2 s i @ T1 R ln
p
i ,1
For constant specific heats,
T2
p2
s c P ln
R ln
T
p
1
1
T
p
w ith s i c P , i ln 2 R ln 2
T1
p1
or
T
V
s cV ln 2 R ln 2
T1
V1
w ith
T
V
s i cV , i ln 2 R ln 2
T1
V1
yi p2
s i @ T 2 s i @ T1 R ln
y
p
i 1
13
Example (12.17): A mixture of 2 kg of H2 and 4 kg of N2 is compressed in a
piston-cylinder assembly in a polytropic process for which n = 1.2. The
temperature increases from 22 to 150°C. Using constant values for the
specific heats, determine
(a) The heat transfer, in kJ.
(b) The entropy change, in kJ/K.
Principles to be applied:
Closed Ideal Gas system:
1st Law of Thermodynamics:
pV nRT
p iV n i R T
and
U U 2 U 1 Q1 2 W 1 2
where for polytropic process
W12
2
2
2nd
Law of thermodynamics:
S S 2 S1
pdV
1
Q
T
1
m R T 2 T1
1 n
14
Example (12.17): A mixture of 2 kg of H2 and 4 kg of N2 is compressed in a
piston-cylinder assembly in a polytropic process for which n = 1.2. The
temperature increases from 22 to 150°C. Using constant values for the
specific heats, determine
(a) The heat transfer, in kJ.
(b) The entropy change, in kJ/K.
Find the moles of each component and the Mixtures Molecular Weight:
nH 2
mH2
M
2 kg
H2
nN2
2 .0 1 8 kg / km o l
0.991km ol
yH2
nH 2
n to ta l
m N2
M
N2
4 kg
2 8 .0 1kg / km o l
0.143km ol
0 .9 9 1km o l
yN2
(0 .9 9 1 0 .1 4 3) km o l
0 .8 7 4
nN2
n to ta l
0 .1 4 3 km o l
(0 .9 9 1 0 .1 4 3) km o l
0.126
Molecular
Weight:
M y H M H y N M N 0.874 2.018 0.126 28.01 5.29
2
2
2
2
15
Example (12.17): A mixture of 2 kg of H2 and 4 kg of N2 is compressed in a
piston-cylinder assembly in a polytropic process for which n = 1.2. The
temperature increases from 22 to 150°C. Using constant values for the
specific heats, determine
(a) The heat transfer, in kJ.
(b) The entropy change, in kJ/K.
To Evaluate work for a polytropic process.
W 12
mR T 2 T1
1 n
m R M
T
2
T1
1 n
8 .3 1 4 kJ / km o l K
2 4 kg
1 5 0 2 2 K
5 .2 9 kg / km o l
1 1 .2
6035.1kJ
16
Example (12.17):
(a) The heat transfer, in kJ.
(b) The entropy change, in kJ/K.
Evaluate change in internal energy
Assuming constant heat capacity from Table A-20 at average temperature (359 K)
U m H 2 c v , H 2 T 2 T1 m N 2 c v , N 2 T 2 T1
2 kg 1 0 .3 1 1 kJ / kg K
1 5 0 2 2 K
4 kg 0 .7 4 5 kJ / kg K
1 5 0 2 2 K
3021.1kJ
Then using the energy balance to evaluate Q
Q1 2 U 1 2 W 1 2
3021.1 6035.1 3014 kJ
17
Example (12.17):
(b) The entropy change, in kJ/K.
T2
V2
s cV ln R ln
T
V
1
1
T2
T2
R
cV ln
ln
T1 n 1 T1
1 n 1
T2
T2
cV ln R ln
T1
T1
R T2
cV
ln
n 1 T1
Using M = 5.29 kg/kmol,
The mixture’s cv (avg. heat capacity) needs to be found.
cV
m f i cV , i m f H 2 cV , H 2 m f N 2 cV , N 2
i
2 kg
4 kg
10.311
kJ
/
kg
K
0.745 kJ / kg K
6 kg
6 kg
so
cV
3.933 kJ
cV
M
cV cV M (3 .9 3 3 kJ / kg K )(5 .2 9 kg / km o l )
20.8 kJ / km ol K
/ kg K
18
Example (12.17):
(b) The entropy change, in kJ/K.
Therefore:
T2
R
s cV
ln
n 1 T1
8.314 kJ / km ol K 423 K
20.8 kJ / km ol K
ln
(1.2 1)
395 K
1 .4 2 2 kJ / km o l K
s
s
M
1 .4 2 2 kJ / km o l K
1 .4 2 2 kJ / km o l K
0 .2 6 9 kJ / kg K
5 .2 9 kg / km o l
S m s (6 kg )( 0 .2 6 9 kJ / kg K ) 1 .6 2 3 kJ / K
Sec 12.4.2 : Mixing of Ideal gases
19
The previous example considered a mixture that had already been formed.
How is a process different when a mixture is formed from two individual
gases which might originally be a different temperatures and pressures?
Whenever two highly ordered substances are mixed, entropy is expected to
increase. This is because
-- Gases are initially at different temperature.
-- Gases are initially at different pressures
-- Gases are distinguishable from one another.
irrev.
process
Sec 12.4.2 : Mixing of Ideal gases
20
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
Start with Energy Balance: Assume no heat lost from system
No change of total volume no work
U Q W 0
where
U 1 n N 2 u N 2 (T N 2 ) n O 2 u O 2 (TO 2 )
U 2 n N 2 u N 2 (T f ) n O 2 u O 2 (T f )
so
or
0 n N 2 u N 2 (T f ) n O 2 u O 2 (T f ) n N 2 u N 2 (T N 2 ) n O 2 u O 2 (T O 2 )
0 n N 2 u N 2 (T f ) u N 2 (T N 2 ) n O 2 u O 2 (T f ) u O 2 (TO 2 )
Assuming constant specific heat, c v _ N
2
and c v _ O 2
0 n N 2 c v _ N 2 (T f T N 2 ) n O 2 c v _ O 2 (T f TO 2 )
Sec 12.4.2 : Mixing of Ideal gases
21
Example 3:
Consider a canister that is initially divided into two equal sized sections. One
side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
Solving for final temperature Tf
Tf
n N 2 c v _ N 2 T N 2 n O 2 c v _ O 2 TO 2
n N 2 cv _ N 2 nO2 cv _ O2
A good choice to use for the temperature to find the cv is the average
temperature of the initial gases (Tave = 400 R). From Table A-20E.
M
c v _ O 2 (@ 400 R ) 0.168 B tu / lb m R
M O 2 32.0 lb m / lb m ol
o
and
so
28.01lb m / lb m ol
c v _ N 2 (@ 400 R ) 0.180 B tu / lb m R
o
cv _ N 2 cv _ N 2 M
N2
N2
(0.180 B tu / lb m R )(28.01lb m / lb m ol ) 5.042 B tu / lb m ol R
c v _ O 2 c v _ O 2 M O 2 (0.168 B tu / lb m R )(32.0 lb m / lb m ol ) 5.376 B tu / lb m ol R
Sec 12.4.2 : Mixing of Ideal gases
22
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
therefore:
Tf
n N 2 c v _ N 2 T N 2 n O 2 c v _ O 2 TO 2
n N 2 cv _ N 2 nO2 cv _ O2
(2 lb m ol )(5.042 B tu / lb m ol R )(500 R ) (3 lb m ol )(5.376 B tu / lb m ol R )(300 R )
(2 lb m ol )(5.042 B tu / lb m ol R ) (3 lb m ol )(5.376 B tu / lb m ol R )
377 R
o
Sec 12.4.2 : Mixing of Ideal gases
23
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
To find the final pressure first find the volume of the total original gases.
V V N 2 VO2
where
p O 2 V O 2 n O 2 R TO 2
p N 2V N 2 n N 2 R TN 2
VN2
n N 2 R TN 2
VO2
pN2
(2 lb m ol )(1.986 B tu / lb m ol R )(500 R ) 1 ft
29.4 lb f / in
365 ft
2
3
778 lbf ft
2
144 in
2
B tu
n O 2 R TO 2
p O2
(3 lb m ol )(1.986 B tu / lb m ol R )(300 R ) 1 ft
14.7 lb f / in
657 ft
total volume is then
V V N 2 V O 2 365 657 1022 ft
3
3
2
778 lbf ft
2
144 in
2
B tu
Sec 12.4.2 : Mixing of Ideal gases
24
Example 3:
Consider a canister that is initially divided into two equal sized sections. One
side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
The mixed gases have a combined Molecular weight given by
M yN2 M
2 lb m ol
5 lb m ol
N2
y O2 M O2
(28.01 lb m / lb m ol )
3 lb m ol
5 lb m ol
(32.00 lb m / lb m ol ) 30.40 lb m / lb m ol
using Ideal Gas equation:
pf
nt R T f
(5 lb m ol )(1.986 B tu / lb m ol R )(377 R ) 778 lb f ft
V
19.79 psi
1022 ft
1.346 atm
3
B tu
1 ft
2
144 in
2
Sec 12.4.2 : Mixing of Ideal gases
25
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
To find the change in entropy:
f
S S f S1
1
where
0
Q
T
S 1 n N 2 s N 2 (T N 2 , p N 2 ) n O 2 s O 2 (TO 2 , p N 2 )
S f n N 2 s N 2 (T f , y N 2 p f ) n O 2 s O 2 (T f , y O 2 p f )
therefore
n N s N (T f , y N p f ) s N (T N , p N ) n O s O (T f , y O p f ) s O (TO , p N )
2
2
2
2
2
2
2
2
2
2
2
2
then using the form based on ideal gas behavior with constant specific heat
nN
2
Tf
c p _ N 2 ln
T
N2
yN p f
2
R ln
p
N2
nO2
Tf
c p _ O 2 ln
T
O2
yO p f
2
R ln
p
O2
Sec 12.4.2 : Mixing of Ideal gases
26
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
To find the change in entropy:
nN
2
Tf
c p _ N 2 ln
T
N2
yN p f
2
R ln
p
N2
nO2
Tf
c p _ O 2 ln
T
O2
yO p f
2
R ln
p
O2
377
(0 .4 )(1 .3 4 6 a tm )
2 lb m o l (5 .0 4 2 B tu / lb m o l R ) ln
(1
.9
8
6
B
tu
/
lb
R
)
ln
m ol
2 a tm
500
377 K
(0 .6 )(1 .3 4 6 a tm )
3 lb m o l ( 4 .3 7 6 B tu / lb m o l R ) ln
(1
.9
8
6
B
tu
/
lb
R
)
ln
m ol
3
0
0
K
1a tm
2.365 4.272
6.637 B tu / R
27
end of slides Lecture 38